AA270 Twin Reverb, bias & phase inverter

[JMadson]

I just finished converting the bias section to AA763 specs, first thing I found was the 470 ohm resistor (before the diode) was WAY to small, I needed more like a 2.5K! Is this normal? In the original curcuit, there's a 1K before the diode, and a 3.3K after so it seems like 2.5K is more like it than 470! Anyway I was wondering how on earth the AB763 curcuit gets you anywhere near the correct bias. All I could get was about 8-10ma per tube witht the stock 470 ohm resistor.
Also, I was wondering about the phase inverter section of the AB763, to me it seems like everyone mods this at the same time as the bias. Does this have some effect on bias? I like the sound of the amp now, (actually, it may be a bit bright), and to me, at least on the schematic, the AA270 looks like more of a balanced design with lower resistor values than the AB763. Do most people simply prefer the the AB763 phase inverter simply because it's "pre-CBS", or does it work better with the bias and output stage?
Thanks in advance for any insight...
Jason

[Enzo]

The 470 ohm is there to protect the diode from surge at power up. The filter cap is where to start things. You can add resustance before the rectifier. nut you can just as easily adjust the voltage divider that follows.

I have AB763 in front of me. There is a 10k pot and a 27k resistor to ground. If you want to lower the bias voltage for higher current through the tubes, lower the value of the 27k. You could just as easily raise the value of the pot, but it is so much easier just to tack in a new resistor.

The AA270 has a 1k before the diode, sure, but it also has a 3.3k above the pot and a 15k below it. That is very different from 10k and 25k. But it is the ratios that matter, not the actual value. Looking at the later drawing I also see that the pot is a balance control, not a bias level control. The 763 is a bias level control, not a balance. SO the pot doesn't enter into the bias level on average in the 270. It is simple enough to rewire it for bias level.

So if all you did was change the 1k to 470, I don't doubt it wasn't right. You have to change the whole voltage divider, not just part. If you want to drive the bias voltage lower in the AA270, reduce the 15k bottom resistor or increase the 3.3k upper resistor in the voltage divider.

Or rewire the bias supply like the 763, and don't forget the power tube grid resistors are also different between the two.

[JMadson]

I did change the to the AB763 bias system. I did the 470 ohm infront of the diode, re-wired the pot to a bias control like in the 763, with the 27K, and changed the two 68K's to 220K like in the 763. Although the lowest the voltage would go was maybe -60 on the wiper of the pot, and that got me about 8-10ma per tube of current. so I should leave the 470ohm, and lower the value of the 27k (maybe to 22K?) to lower the voltage? Changing the 470 ohm to 2.5K gets me the range I need, but I guess I went about it all wrong...
Let me know if I'm on the right track...
Jason

[Enzo]

Well whatever works works, I don't know that wrong is the word.

What is the B+? On AB763 the B+ shows 460v while on the AA270 it shows 410v. This difference also requires changing the bias point. I don't know what the AC is at the bias tap on either amp, but that could be part of it - the raw voltage is too high for the divider values. Note teh 270 wants -44 on the grids while the 763 wants -52 on the grids. That is a substantial difference.

Look at the raw voltage - the DC at the first filter in the bias supply, the one at the end of the diode. That is the voltage you have to work with - the highest bias voltage available in the amp. Now you can calculate the values of the resistors by looking at it as a voltage divider. While the resistor before the diode can make a difference, it is not necessary to change it. Look at AA270. The raw voltage is -52v right out of the diode. Then there is a 3.3k and a 15k resistor to ground - total of 18.3k. Voltage divider resistors have voltage drops proportional to their portion of the total resistance.

So in this example, the voltage across the 15k resistor will be 15k/18.3k times the -52 volts. That means 15/18.3 x 52 = 42.6. Fender says 44 or about a volt different. suspect that was either a round off or a measurement that ignored variation in the parts. In any case, close enough.

Whatever bias voltage you need, start with whatever raw voltage you have, then calculate what resistances you need to get the bias range where you want it. With the 27k, lower it to whatever works. 22k is a good guess for starts.

[JMadson]

Well, I think I'm on the right track now, I've got around 68 VAC out the bias tap. I think I'll go back to the 1K before the diode, and then see what voltage I have at the negative end of the bias cap. I did change the 68K resistors that actually do the "splitting" to 220K as per the 763, do you think I should shoot for -52 volts instead of -44 on the wiper since the two resistors have larger value?
Thanks again...
Jason

[JMadson]

Got it!

Turns out a 1K between the diode and transformer, and a 15K on the pot, like in the AA270, just no 3.3K from the cap to the pot center tap, that and 220K's to split to the two sides of the power tubes, and we're right on the money, I can adjust from -43 to -68 volts on the wiper of the pot now, that translates to about 1-50 ma of dissapation with the bias probe.
Thanks for your help, Enzo...
Jason

[Enzo]

It is just a matter of setting the voltage divider up right. Glad you got it.

I just coverd voltage dividers with my local learner. And in fact one example i was using for the lesson was bias supplies.